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29 Jun 2012 03:31:57 IST

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law of malus

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could u explain law of malus

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shashank shekhar

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30 Jun 2012 21:41:29 IST

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According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer.

i.e I ∞ cos²θ

Suppose the angle between the transmission axes of the analyzer and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyzer. If E₀ is the amplitude of the electric vector transmitted by the polarizer, then intensity I₀ of the light incident on the analyzer is
I ∞ E₀²

The electric field vector E₀ can be resolved into two rectangular components i.e E₀ cosθ and E₀ sinθ. The analyzer will transmit only the component ( i.e E₀ cosθ ) which is parallel to its transmission axis. However, the component E₀sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,

I ∞ ( E₀ x cosθ )²

I / I₀ = ( E₀ x cosθ )² / E₀² = cos²θ

I = I₀ x cos²θ

Therefore, I ∞ cos²θ. This proves law of malus.

When θ = 0° ( or 180° ), I = I₀ cos²0° = I₀ That is the intensity of light transmitted by the analyzer is maximum when the transmission axes of the analyzer and the polarizer are parallel.

When θ = 90°, I = I₀ cos²90° = 0 That is the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other.

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